Question

Question: Consider the situation described in Exercise 7 of Sec. 8.5. Use a prior distribution from the normal-gamma family with values \({{\bf{\mu }}_{\bf{0}}}{\bf{ = 150,}}{{\bf{\lambda }}_{\bf{0}}}{\bf{ = 0}}{\bf{.5,}}{{\bf{\alpha }}_{\bf{0}}}{\bf{ = 1,}}{{\bf{\beta }}_{\bf{0}}}{\bf{ = 4}}\)

a. Find the posterior distribution of \({\bf{\mu }}\,\,{\bf{and}}\,\,{\bf{\tau = }}\frac{{\bf{1}}}{{{{\bf{\sigma }}^{\bf{2}}}}}\)

b. Find an interval (a, b) such that the posterior probability is 0.90 that a <\({\bf{\mu }}\)<b.

Short answer

(a) Normal-gamma with hyperparameters \({\mu _1} = 156.7,{\lambda _1} = 20.5,{\alpha _1} = 11,{\beta _1} = 4885.7\)

(b) (148.7, 164.7).

Step-by-step solution

Given information

It is given that two variables \(\mu \,\,{\rm{and}}\,\,\tau \)have the joint normal-gamma distribution such that \({\mu _0} = 150,{\lambda _0} = 0.5,{\alpha _0} = 1,{\beta _0} = 4\).

The points described in Exercise 7 of Sec. 8.5 are: 186, 181, 176, 149, 184, 190, 158, 139, 175, 148, 152, 111, 141, 153, 190, 157, 131, 149, 135, 132.

Define Normal-Gamma distribution

Let \(\mu \,\,{\rm{and}}\,\,\tau \) be random variables. Suppose that the conditional distribution of \(\mu \,\,{\rm{given}}\,\,\tau \) is the normal distribution with mean \({\mu _0}\) and precision \({\lambda _0}\tau \) .

Suppose also that the marginal distribution of \(\,\tau \) is the gamma distribution with parameters \({\alpha _0}\,\,{\rm{and}}\,\,{\beta _0}\).

Then we say that the joint distribution of\(\mu \,\,and\,\,\tau \) is the normal-gamma distribution with hyperparameters \({\mu _0},{\lambda _0},{\alpha _0},{\beta _0}\).

Define the posterior variables

We know that,

\(\begin{aligned}{}{\mu _1} &= \frac{{{\lambda _0}{\mu _0} + n\overline {{x_n}} }}{{{\lambda _0} + n}}\\{\lambda _1} &= {\lambda _0} + n\\{\alpha _1} &= {\alpha _0} + \frac{n}{2}\\{\beta _1} &= {\beta _0} + \frac{{{s_n}^2}}{2} + \frac{{n{\lambda _0}{{\left( {\overline {{x_n}} - {\mu _0}} \right)}^2}}}{{2\left( {{\lambda _0} + n} \right)}}\end{aligned}\)

Also, the values of:

\(\begin{aligned}{}\overline {{x_n}} &= \frac{{186 + 181 + {\rm{ }} \ldots + {\rm{ }}135 + {\rm{ }}132}}{{20}}\\ &= 156.85\end{aligned}\)

\(\begin{aligned}{}{s_n}^2 &= \frac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline {{x_n}} } \right)}^2}} }}{{n - 1}}\\ &= \frac{{5315.525}}{{19}}\\ &= 279.7645\end{aligned}\)

(a) Substitute the values

\(\begin{aligned}{}{\mu _1} &= \frac{{150 \times 0.5 + 20 \times 156.85}}{{1 + 20}}\\ &= 156.7\\{\lambda _1} &= 0.5 + 20\\ &= 20.5\end{aligned}\)

\(\begin{aligned}{}{\alpha _1} &= 1 + \frac{{20}}{2}\\ &= 11\\{\beta _1} &= 4 + \frac{{279.76}}{2} + \frac{{20 \times 0.5 \times {{\left( {156.85 - 150} \right)}^2}}}{{2\left( {0.5 + 20} \right)}}\\ &= 4885.7\end{aligned}\)

Therefore, the answer is: \({\mu _1} = 156.7,{\lambda _1} = 20.5,{\alpha _1} = 11,{\beta _1} = 4885.7\)

(b) Define a new variable

Let,

\(\begin{aligned}{}U &= {\left( {\frac{{{\lambda _1}{\alpha _1}}}{{{\beta _1}}}} \right)^{\frac{1}{2}}}\left( {\mu - {\mu _1}} \right)\\ &= 0.5939\left( {\mu - 156.7} \right)\end{aligned}\)

Here U follows t distribution with \(2{\alpha _1}\) degrees of freedom, that is 22.

Now,

\(\begin{aligned}{}P\left( {a < \mu < b} \right) &= 0.90\\ \Rightarrow P\left( {0.5939\left( {a - 156.7} \right) < 0.5939\left( {\mu - 156.7} \right) < 0.5939\left( {b - 156.7} \right)} \right) &= 0.90\\ \Rightarrow P\left( {0.5939\left( {a - 156.7} \right) < U < 0.5939\left( {b - 156.7} \right)} \right) &= 0.90\end{aligned}\)

The confidence intervals derived for U are -1.71714and 1.71714

Therefore, the value of a and b is,

\(\begin{aligned}{}0.5939\left( {a - 156.7} \right) &= - {\rm{1}}{\rm{.71714}}\\ \Rightarrow a &= 148.7\end{aligned}\)

And,

\(\begin{aligned}{}0.5939\left( {b - 156.7} \right) &= {\rm{1}}{\rm{.71714}}\\ \Rightarrow b &= 164.7\end{aligned}\)

Therefore, the interval is (148.7, 164.7).